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ps3-solution - Econ 805 Problem Set 3 Solution 1(Gibbons...

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Econ 805 Problem Set 3 Solution 1. (Gibbons 1.2). In the following game, what strategies survive IESDS (iterated elimination of strictly dominated strategies)? What are the pure strategy Nash equilibria? L C R T 2 , 0 1 , 1 4 , 2 M 3 , 4 1 , 2 2 , 3 B 1 , 3 0 , 2 3 , 0 Answer: for player 1, B is strictly dominated by T . So we can delete the row of B . Then for player 2, C is strictly dominated by R . No further strategies can be deleted. Therefore, the strategies that survive IESDS are: ( T, M ) and ( L, R ) . The pure strategy NE of this game are ( M, L ) and ( T, R ) (see the underscore in the bi-matrix). 2. Mascolell 8.B.5. Answer: (i) First, any quantity q > q m = a c 2 b (the monopoly quantity) is strictly dominated by q m . Then the remaining strategy set for each fi rm is [0 , q m ] . Now you can calculate the best response to q m . You will get R ( q m ) = a c 2 b q m 2 .. One can show that any q < b ( q m ) is strictly dominated by R ( q m ) . Now the relevant strategy set is [ R ( q m ) , q m ] . Now you can calculate the best response to R ( q m ) . You can easily get that R 2 ( q m ) = a c 2 b R ( q m ) 2 . Again you can prove that any quantity q > R 2 ( q m ) is strictly dominated by R 2 ( q m ) . Now the relevant strategy set shrinks to [ R ( q m ) , R 2 ( q m )] . If you continue this process, after 2 n rounds of IESDS, the remaining strategy set is [ R 2 n 1 ( q m ) , R 2 n ( q m )] . Now you only need to show that R n ( q m ) converges as n goes to in fi nity. You can show that R n ( q m ) = 2 n 2 n 1 + 2 n 2 + ... + ( 1) n 2 n n 2 n q m = [1 1 2 + 1 4 ... + ( 1) n 1 2 n ] q m = ( { 1 1 4 [1 + 1 4 + ... + 1 4 n 2 1 ] } q m if n is even 1 2 [1 + 1 4 + .. + 1 4 n 1 2 ] q m if n is odd If n is even, R n ( q m ) = { 1 1 4 1 ( 1 4 ) n 2 1 1 1 4 } q m 2 3 q m If n is odd, R n ( q m ) = 1 2 1 ( 1 4 ) n 1
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